![HCMUT / 2020
DungTrinh, PhD
Dept. of Telecoms Engineering
5a. Propagation Constant andVelocity
24
❖ Propagation constant:
𝛼: attenuation constant [Np/m] or [dB/m].
𝛽: phase constant [rad/s].
❖ The “wave velocity” is described by its “phase velocity”. Since velocity is
change in distance with respect to time, we need to first express the
propagation wave in its real form:
❖ Let’s set the absolute phase to some arbitrary value: 𝝎𝒕 − 𝜷𝒛 = 𝝓𝒄. Then:
and
𝜸 𝝎 = 𝜶 𝝎 + 𝒋𝜷 𝝎 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪
𝑽+
𝒛, 𝒕 = 𝕽𝒆 𝑽+
𝒛 𝒆−𝒋𝝎𝒕 = 𝑽𝟎
+
𝒄𝒐𝒔 𝝎𝒕 − 𝜷𝒛
𝒛 =
𝝎𝒕 − 𝝓𝒄
𝜷
𝒗𝒑 =
𝝏𝒛
𝝏𝒕
=
𝝎
𝜷
𝛼 𝑑𝐵/𝑚 = 20𝑙𝑜𝑔10𝑒𝛼 𝑁𝑝/𝑚 = 8.68𝛼 𝑁𝑝/𝑚](https://image.slidesharecdn.com/chapter12024-250316183117-5fb71144/85/Chapter-1_2024-pdf-Microwave-Engineering-24-320.jpg)
Chapter 1_2024.pdf Microwave Engineering
- 1. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Microwave Engineering 1 M.Eng. Dang Ngoc Hanh hanhdn@hcmut.edu.vn
- 2. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 2 Trinh Xuan Dung, PhD dung.trinh@hcmut.edu.vn Department of Telecommunications Faculty of Electrical and Electronics Engineering Ho Chi Minh city University of Technology Chapter 1 Theory and Applications of Transmission Lines
- 3. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Contents 3 1. Introduction 2. Lumped-Element Circuit Model for Transmission Lines 3. Transmission Line Equations and Solutions 4. Characteristic Impedance of Transmission Line 5. Propagation constant and velocity 6. Lossless and Lossy Transmission Lines 7. Reflection Coefficient 8. Transmission Line Impedance and Admittance 9. Power Transmission on Transmission Lines 10. Standing Wave and Standing Wave Ratio 11. Impedance Matching Problems
- 4. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 1. Introduction 4 ❖ The previous class provided the analysis of EM field and wave traveling in the free space. This chapter provides the analysis of wave propagations in the guided mediums : transmission lines. ❖ For efficient point-to-point transmission of power and information, the source energy must be directed or guided. ❖ The key difference between circuit theory and Transmission Line is electrical size. ❖ At low frequencies, an electrical circuit is completely characterized by the electrical parameters like resistance, inductance, capacitance etc. and the physical size of the electrical components plays no role in the circuit analysis. ❖ As the frequency increases however, the size of the components becomes important. The voltage and currents exist in the form of waves. Even a change in the length of a simple connecting wire may alter the behavior of the circuit.
- 5. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 1. Introduction 5 ❖ The circuit approach then has to be re-investigated with inclusion of the space into the analysis. This approach is then called the Transmission Line approach. ❖ Although the primary objective of a transmission line is to carry electromagnetic energy efficiently from one location to other, they find wide applications in high frequency circuit design. ❖ Also at high frequencies, the transmit time of the signals can not be ignored. In the era of high speed computers, where data rates are approaching to few Gb/sec, the phenomena related to the electromagnetic waves, like the bit distortion, signal reflection, impedance matching play a vital role in high speed communication networks.
- 6. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 1. Introduction 6 At a given location along the line, find: ❖ Current, voltage and power ❖ Reflection coefficient, impedance, VSWR ❖ Design real TLs, such as micro-strip lines, CPW lines General problems of the chapter ZL ZS VS I(l) V(l) z l 0 Characteristic Impedance Z0
- 7. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 2. Lumped-Element Circuit Model forTransmission Lines 7 Examples of Transmission Lines: Two-wire TL Coaxial TL Microstrip TL ❖ Two-wire Transmission Line: consists of a pair of parallel conducting wires separated by a uniform distance. Examples: telephone line, cable connecting from roof-top antenna to TV receiver. ❖ Coaxial Transmission Line: consists of inner conductor and and a coaxial outer separated by a dielectric medium. Examples: TV Cable, etc. ❖ Microstrip Transmission Line: consists of two parallel conducting plates separated by a dielectric slab. It can be fabricated inexpensively on PCB.
- 8. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 2. Lumped-Element Circuit Model forTransmission Lines 8 ZL ZS VS I(l) V(l) z l 0 ❖ Current i and voltage v are a function of position z because a wire is never a “perfect” conductor. It will have: ▪ Inductance (G) ▪ Resistance (R) ▪ Capacitance (C) ▪ Conductance (L)
- 9. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 2. Lumped-Element Circuit Model forTransmission Lines 9 R, L, G, and C are per-unit-length quantities defined as follows: ❖ R = series resistance per unit length, for both conductors, in /m. ❖ L = series inductance per unit length, for both conductors, in H/m. ❖ G = shunt conductance per unit length, in S/m. ❖ C = shunt capacitance per unit length, in F/m. o Series inductance L represents the total self-inductance of the two conductors. o Shunt capacitance C is due to the close proximity of the two conductors. o Series resistance R represents the resistance due to the finite conductivity of the individual conductors. o Shunt conductance G is due to dielectric loss in the material between the conductors. o R and G, therefore, represent loss.
- 10. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 2. Lumped-Element Circuit Model forTransmission Lines 10 Table: Transmission Line Parameters of some common lines: Further reading: Kỹ thuật SCT, p.25-p.33
- 11. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 11 Applying Kirchoff’s Voltage Law (KVL): Applying Kirchoff’s Current Law (KCL): 𝒗 𝒛 + 𝚫𝒛, 𝒕 = 𝒗 𝒛, 𝒕 − 𝑹𝚫𝒛𝒊 𝒛, 𝒕 − 𝑳𝚫𝒛 𝝏𝒊 𝒛, 𝒕 𝝏𝒕 𝒊 𝒛 + 𝚫𝒛, 𝒕 = 𝒊 𝒛, 𝒕 − 𝑮𝚫𝒛𝒗 𝒛, 𝒕 − 𝑪𝚫𝒛 𝝏𝒗 𝒛, 𝒕 𝝏𝒕
- 12. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 12 Then: 𝒗 𝒛 + 𝚫𝒛, 𝒕 − 𝒗 𝒛, 𝒕 𝚫𝒛 = −𝑹𝒊 𝒛, 𝒕 − 𝑳 𝝏𝒊 𝒛, 𝒕 𝝏𝒕 𝒊 𝒛 + 𝚫𝒛, 𝒕 − 𝒊 𝒛, 𝒕 𝚫𝒛 = −𝑮𝒗 𝒛, 𝒕 − 𝑪 𝝏𝒗 𝒛, 𝒕 𝝏𝒕 When ∆𝑧 → 0: 𝝏𝒗(𝒛, 𝒕) 𝝏𝒛 = −𝑹𝒊 𝒛, 𝒕 − 𝑳 𝝏𝒊 𝒛, 𝒕 𝝏𝒕 𝝏𝒊(𝒛, 𝒕) 𝝏𝒛 = −𝑮𝒗 𝒛, 𝒕 − 𝑪 𝝏𝒗 𝒛, 𝒕 𝝏𝒕 These equations are “telegrapher’s equations”. There are infinite number of solutions 𝒗 𝒛, 𝒕 and 𝒗 𝒛, 𝒕 for the “telegrapher’s equations”. The problem can be simplified by assuming that the function of time is “time harmonic” (sinusoidal). 𝒊 𝒕 = 𝑰𝑷𝒄𝒐𝒔(𝝎𝒕) 𝐼𝑃 = 𝑰𝑷 = 𝑽𝑷 𝒁𝑪 𝒊𝒄 𝒕 = −𝑪 𝜹𝒗𝒄(𝒕) 𝜹𝒕
- 13. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 13 ❖ If a sinusoidal voltage source with frequency 𝜔 is used to excite a linear, time-invariant circuit then the voltage at every point with the circuit will likewise vary sinusoidal. ❖ The voltage along a transmission line when excited by a sinusoidal source must have the form: ❖ The time harmonic voltage at every location z along a transmission line: where: and ❖ There is no reason to explicitly write the complex function 𝒆𝒋𝝎𝒕 since the only unknown is the complex function 𝑽 𝒛 . Once we determine 𝑽 𝒛 , we can always recover the real function 𝒗 𝒛, 𝒕 : 𝒗 𝒛, 𝒕 = 𝒗 𝒛 𝒄𝒐𝒔 𝝎𝒕 + 𝝋 𝒛 = 𝕽𝒆 𝒗 𝒛 𝒆𝒋𝝎𝒕𝒆𝒋𝝋 𝒛 𝑽 𝒛 = 𝒗 𝒛 𝒆𝒋𝝋 𝒛 𝒗 𝒛 = 𝑽 𝒛 𝝋 𝒛 = 𝒂𝒓𝒈 𝑽 𝒛 𝒗 𝒛, 𝒕 = 𝕽𝒆 𝑽 𝒛 𝒆𝒋𝝎𝒕
- 14. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 14 ❖ Let’s assume that 𝒗 𝒛, 𝒕 and 𝒊 𝒛, 𝒕 each have the time harmonic form: ❖ Then time derivative of these functions are: ❖ The telegrapher’s equations thus become: 𝒗 𝒛, 𝒕 = 𝕽𝒆 𝑽 𝒛 𝒆𝒋𝝎𝒕 𝒊 𝒛, 𝒕 = 𝕽𝒆 𝑰 𝒛 𝒆𝒋𝝎𝒕 𝝏𝒗(𝒛, 𝒕) 𝝏𝒛 = 𝕽𝒆 𝒋𝝎𝑽 𝒛 𝒆𝒋𝝎𝒕 𝝏𝒊(𝒛, 𝒕) 𝝏𝒛 = 𝕽𝒆 𝒋𝝎𝑰 𝒛 𝒆𝒋𝝎𝒕 𝕽𝒆 𝝏𝑽 𝒛 𝝏𝒛 𝒆𝒋𝝎𝒕 = 𝕽𝒆 − 𝑹 + 𝒋𝝎𝑳 𝑰 𝒛 𝒆𝒋𝝎𝒕 𝕽𝒆 𝝏𝑰 𝒛 𝝏𝒛 𝒆𝒋𝝎𝒕 = 𝕽𝒆 − 𝑮 + 𝒋𝝎𝑪 𝑽 𝒛 𝒆𝒋𝝎𝒕
- 15. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 15 𝝏𝑽 𝒛 𝝏𝒛 = − 𝑹 + 𝒋𝝎𝑳 𝑰 𝒛 𝝏𝑰 𝒛 𝝏𝒛 = − 𝑮 + 𝒋𝝎𝑪 𝑽 𝒛 ❖ Then the complex form of telegrapher’s equations are: Note that these functions are not a function of time t. ❖ Take the derivative with respect to z of the telegrapher’s equations, lead to: Complex Value: 𝒗 𝒛 𝒆𝒋𝝋 𝒛 𝝏𝟐𝑽 𝒛 𝝏𝒛𝟐 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪 𝑽 𝒛 𝝏𝟐𝑰 𝒛 𝝏𝒛𝟐 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪 𝑰 𝒛
- 16. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 16 ❖ These equations can be written as: where 𝜸 𝝎 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪 is propagation constant. ❖ Only special equations satisfy these equations. The solution of these equations can be found as: where 𝜸 = 𝜶 + 𝒋𝜷. 𝝏𝟐𝑽 𝒛 𝝏𝒛𝟐 = 𝜸𝟐 𝝎 𝑽 𝒛 𝝏𝟐 𝑰 𝒛 𝝏𝒛𝟐 = 𝜸𝟐 𝝎 𝑰 𝒛 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆+𝜸𝒛
- 17. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 17 ❖ The time harmonic voltage at every location z along a transmission line: where: and ❖ Only special equations satisfy these equations. The solution of these equations can be found as: where 𝜸 = 𝜶 + 𝒋𝜷. 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆+𝜸𝒛 𝑽 𝒛 = 𝒗 𝒛 𝒆𝒋𝝋 𝒛 𝒗 𝒛 = 𝑽 𝒛 𝝋 𝒛 = 𝒂𝒓𝒈 𝑽 𝒛
- 18. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 3.Transmission Line Equations and Solution 18 ❖ The current and voltage at a given point must have the form: 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜶𝒛 𝒆−𝒋𝜷𝒛 + 𝑽𝟎 − 𝒆+𝜶𝒛 𝒆+𝒋𝜷𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜶𝒛𝒆−𝒋𝜷𝒛 + 𝑰𝟎 − 𝒆+𝜶𝒛𝒆+𝒋𝜷𝒛 ZL ZS VS I(l) V(l) z 0 Incident wave Reflected wave 𝑽𝟎 + 𝑰𝟎 + 𝜸 𝝎 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪 is propagation constant. 𝜸 = 𝜶 + 𝒋𝜷
- 19. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Summary 1 19 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜶𝒛𝒆−𝒋𝜷𝒛 + 𝑽𝟎 − 𝒆+𝜶𝒛𝒆+𝒋𝜷𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜶𝒛𝒆−𝒋𝜷𝒛 + 𝑰𝟎 − 𝒆+𝜶𝒛𝒆+𝒋𝜷𝒛 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆+𝜸𝒛 𝜸 𝝎 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪 is propagation constant. 𝜸 = 𝜶 + 𝒋𝜷
- 20. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Q&A 20
- 21. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 21 ZL ZS VS I(l) V(l) z 0 Incident wave Reflected wave
- 22. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 4. Characteristic Impedance ofTransmission Line 22 ❖ The terms in each equation describe two waves propagating in the transmission line, one propagating in one direction (+z) and the other wave propagating in the opposite direction (-z): ❖ Then: ❖ After re-arranging, 𝑰 𝒛 must be: ❖ For the equations to be true for all z, I0 and V0 must be related as: 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆+𝜸𝒛 𝝏𝑽 𝒛 𝝏𝒛 = −𝜸𝑽𝟎 + 𝒆−𝜸𝒛 + 𝜸𝑽𝟎 − 𝒆+𝜸𝒛 = − 𝑹 + 𝒋𝝎𝑳 𝑰 𝒛 𝑰 𝒛 = 𝜸 𝑹 + 𝒋𝝎𝑳 𝑽𝟎 + 𝒆−𝜸𝒛 − 𝜸 𝑹 + 𝒋𝝎𝑳 𝑽𝟎 − 𝒆+𝜸𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆−𝜸𝒛 𝑰𝟎 + = 𝑽𝟎 + 𝒁𝟎 𝑰𝟎 − = 𝑽𝟎 − 𝒁𝟎 and where: 𝒁𝟎 = 𝑹 + 𝒋𝝎𝑳 𝜸 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪
- 23. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 4. Characteristic Impedance ofTransmission Line 23 ❖ 𝑽𝟎 + and 𝑰𝟎 + are determined by the “boundary condition” (what is connected to either end of the transmission line) but the ratio 𝑽𝟎 + 𝑰𝟎 + is determined by the parameters of the transmission line only. ❖ Set 𝑍 = 𝑅 + 𝑗𝜔𝐿 and 𝑌 = 𝐺 + 𝑗𝜔𝐶. Then: ❖ Lossless transmission line: ❖ In practice: ❖ 𝒁𝟎 is always real. ❖ In communications system: 𝒁𝟎 = 𝟓𝟎𝜴. In telecommunications: : 𝒁𝟎 = 𝟕𝟓𝜴. 𝒁𝟎 = 𝑍Δ𝑥 + 1 𝑌Δ𝑥 ∥ 𝑍0 𝑥→0 𝑍 𝑌 = 𝑅 + 𝑗𝜔𝐿 𝐺 + 𝑗𝜔𝐶 𝒁𝟎 = 𝐿 𝐶
- 24. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 5a. Propagation Constant andVelocity 24 ❖ Propagation constant: 𝛼: attenuation constant [Np/m] or [dB/m]. 𝛽: phase constant [rad/s]. ❖ The “wave velocity” is described by its “phase velocity”. Since velocity is change in distance with respect to time, we need to first express the propagation wave in its real form: ❖ Let’s set the absolute phase to some arbitrary value: 𝝎𝒕 − 𝜷𝒛 = 𝝓𝒄. Then: and 𝜸 𝝎 = 𝜶 𝝎 + 𝒋𝜷 𝝎 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪 𝑽+ 𝒛, 𝒕 = 𝕽𝒆 𝑽+ 𝒛 𝒆−𝒋𝝎𝒕 = 𝑽𝟎 + 𝒄𝒐𝒔 𝝎𝒕 − 𝜷𝒛 𝒛 = 𝝎𝒕 − 𝝓𝒄 𝜷 𝒗𝒑 = 𝝏𝒛 𝝏𝒕 = 𝝎 𝜷 𝛼 𝑑𝐵/𝑚 = 20𝑙𝑜𝑔10𝑒𝛼 𝑁𝑝/𝑚 = 8.68𝛼 𝑁𝑝/𝑚
- 25. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 25 5b. Line Impedance ❖ The Line Impedance is NOT the T.L Impedance 𝒁𝟎. Recall that: ❖ Therefore, the Line Impedance can be written as: ❖ Or more specifically: 𝑽 𝒛 = 𝑽+ 𝒛 + 𝑽− 𝒛 𝑰 𝒛 = 𝑽+ 𝒛 − 𝑽− 𝒛 𝒁𝟎 𝒁 𝒛 = 𝑽 𝒛 𝑰 𝒛 = 𝒁𝟎 𝑽+ 𝒛 + 𝑽− 𝒛 𝑽+ 𝒛 − 𝑽− 𝒛 𝒁 𝒛 = 𝒁𝟎 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆𝜸𝒛 𝑽𝟎 + 𝒆−𝜸𝒛 − 𝑽𝟎 − 𝒆𝜸𝒛
- 26. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 6. Lossless and Low-lossTransmission Line 26 ❖ In practice, transmission lines have losses due to finite conductivity and/or lossy dielectric but these losses are usually small. ❖ In most practical microwave: ▪ Losses may be neglected → Lossless Transmission Line. ▪ Losses may be assumed to be very small → Low-loss Transmission Line. ❖ Lossless Transmission Line: 𝑹 = 𝟎, 𝑮 = 𝟎 ❖ Low-loss Transmission Line: both conductor and dielectric loss will be small, and we can assume that 𝑅 ≪ 𝜔𝐿 and 𝐺 ≪ 𝜔𝐶. Then: 𝑅𝐺 ≪ 𝜔2𝐿𝐶. Then: 𝜸 𝝎 = 𝑹 + 𝒋𝝎𝑳 𝑮 + 𝒋𝝎𝑪 = 𝒋𝝎 𝑳𝑪 𝜶 𝝎 = 𝟎 𝜷 𝝎 = 𝝎 𝑳𝑪 𝜸 𝝎 ≃ 𝒋𝝎 𝑳𝑪 𝟏 − 𝒋 𝑹 𝝎𝑳 + 𝑮 𝝎𝑪
- 27. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 6. Lossless and Low-lossTransmission Line 27 ❖ Using the Taylor series expansion* for: ❖ Then: ❖ Hence: where: 𝑍0 = 𝑅+𝑗𝜔𝐿 𝐺+𝑗𝜔𝐶 ≃ 𝐿 𝐶 * https://en.wikipedia.org/wiki/Taylor_series 𝜸 𝝎 ≃ 𝒋𝝎 𝑳𝑪 𝟏 − 𝒋 𝑹 𝝎𝑳 + 𝑮 𝝎𝑪 ≃ 𝒋𝝎 𝑳𝑪 𝟏 − 𝒋 𝟐 𝑹 𝝎𝑳 + 𝑮 𝝎𝑪 𝟏 + 𝒙 ≃ 𝟏 + Τ 𝒙 𝟐 − ൗ 𝒙𝟐 𝟖 + ൗ 𝒙𝟑 𝟏𝟔 + ⋯ 𝜶 ≃ 𝟏 𝟐 𝑹 𝑪 𝑳 + 𝑮 𝑳 𝑪 = 𝟏 𝟐 𝑹 𝒁𝟎 + 𝑮𝒁𝟎 𝜷 ≃ 𝝎 𝑳𝑪
- 28. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 7. Reflection Coefficient 28 ❖ Voltage Reflection Coefficient is defined as: ❖ Current Reflection Coefficient is defined as: 𝚪𝑽 𝒛 = 𝑹𝒆𝒇𝒍𝒆𝒄𝒕𝒆𝒅 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 𝑰𝒏𝒄𝒊𝒅𝒆𝒏𝒕 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 = 𝑽𝟎 − 𝒆+𝜸𝒛 𝑽𝟎 + 𝒆−𝜸𝒛 = 𝑽𝟎 − 𝑽𝟎 + 𝒆𝟐𝜸𝒛 𝒁𝑳 𝒁𝑺 𝑽𝑺 𝑰(𝒍) 𝑽(𝒍) 𝒛 𝒍 𝟎 Incident wave Reflected wave 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆+𝜸𝒛 𝚪𝑰 𝒛 = 𝑹𝒆𝒇𝒍𝒆𝒄𝒕𝒆𝒅 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝑰𝒏𝒄𝒊𝒅𝒆𝒏𝒕 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 = 𝑰𝟎 − 𝒆+𝜸𝒛 𝑰𝟎 + 𝒆−𝜸𝒛 = − ൗ 𝑽𝟎 − 𝒁𝟎 ൘ 𝑽𝟎 + 𝒁𝟎 𝒆𝟐𝜸𝒛 = −𝚪𝑽 𝒛
- 29. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 7. Reflection Coefficient 29 ❖ At load: ❖ Note that: ❖ Then: ❖ At location z: 𝚪𝑳 = 𝑽𝟎 − 𝑽𝟎 + 𝒆𝟐𝜸𝒍 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆+𝜸𝒛 𝒁𝑳 = 𝑽(𝒍) 𝑰(𝒍) = 𝒁𝟎 𝑽𝟎 + 𝒆−𝒋𝜷𝒍 + 𝑽𝟎 − 𝒆𝒋𝜷𝒍 𝑽𝟎 + 𝒆−𝒋𝜷𝒍 − 𝑽𝟎 − 𝒆𝒋𝜷𝒍 = 𝒁𝟎 𝟏 + 𝚪𝑳 𝟏 − 𝚪𝑳 𝚪𝑳 = 𝒁𝑳 − 𝒁𝟎 𝒁𝑳 + 𝒁𝟎 𝒁𝑳 𝒁𝑺 𝑽𝑺 𝑰(𝒍) 𝑽(𝒍) 𝒛 𝒍 0 Incident wave Reflected wave 𝒛 = 𝒍 − 𝒅 𝚪 𝒛 = 𝒍 − 𝒅 = 𝑽𝟎 − 𝑽𝟎 + 𝒆𝟐𝜸𝒛 = 𝑽𝟎 − 𝑽𝟎 + 𝒆𝟐𝜸 𝒍−𝒅 = 𝑽𝟎 − 𝑽𝟎 + 𝒆𝟐𝜸𝒍𝒆−𝟐𝜸𝒅 = 𝚪𝑳𝒆−𝟐𝜸𝒅
- 30. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 7. Reflection Coefficient - Representation on a complex plane 30 ❖ Reflection Coefficient at 𝑧 = 𝑙 − 𝑑: where: 𝛾 = 𝛼 + 𝑗𝛽. ❖ Then: 𝚪 𝒛 = 𝒍 − 𝒅 = 𝚪𝑳𝒆−𝟐𝜸𝒅 𝚪 𝒛 = 𝒍 − 𝒅 = 𝚪𝑳𝒆−𝟐𝜶𝒅 𝒆−𝟐𝒋𝜷𝒅 / 2 2 2 2 2 2 2 2 = = = = d d d
- 31. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 31 8.Transmission Line Impedance andAdmittance 𝒁(𝒙) ❖ The line impedance at 𝑧 = 𝑙 − 𝑑: ❖ Note that: ❖ Then the line impedance can be specified: ❖ More specifically: 𝒁 𝒛 = 𝒁𝟎 𝑽𝟎 + 𝒆−𝒋𝜸𝒛 + 𝑽𝟎 − 𝒆𝒋𝜸𝒛 𝑽𝟎 + 𝒆−𝒋𝜸𝒛 − 𝑽𝟎 − 𝒆𝒋𝜸𝒛 𝚪 𝒛 = 𝒍 − 𝒅 = 𝑽𝟎 − 𝑽𝟎 + 𝒆𝟐𝜸𝒍𝒆−𝟐𝜸𝒅 = 𝚪𝑳𝒆−𝟐𝜸𝒅 𝒁 𝒛 = 𝒁𝟎 𝟏 + 𝚪 𝒛 𝟏 − 𝚪 𝒛 𝒁 𝒛 = 𝒁𝟎 𝒁𝑳 + 𝒁𝟎 𝒆𝜸𝒅 + 𝒁𝑳 − 𝒁𝟎 𝒆−𝜸𝒅 𝒁𝑳 + 𝒁𝟎 𝒆𝜸𝒅 − 𝒁𝑳 − 𝒁𝟎 𝒆−𝜸𝒅 = 𝒁𝟎 𝒁𝑳 𝒆𝜸𝒅 + 𝒆−𝜸𝒅 + 𝒁𝟎 𝒆𝜸𝒅 − 𝒆−𝜸𝒅 𝒁𝑳 𝒆𝜸𝒅 − 𝒆−𝜸𝒅 + 𝒁𝟎 𝒆𝜸𝒅 + 𝒆−𝜸𝒅 = 𝒁𝟎 𝒁𝑳𝒄𝒐𝒔𝒉 𝜸𝒅 + 𝒁𝟎𝒔𝒊𝒏𝒉 𝜸𝒅 𝒁𝑳𝒔𝒊𝒏𝒉 𝜸𝒅 + 𝒁𝟎𝒄𝒐𝒔𝒉 𝜸𝒅 = 𝒁𝟎 𝒁𝑳 + 𝒁𝟎𝒕𝒂𝒏𝒉 𝜸𝒅 𝒁𝟎 + 𝒁𝑳𝒕𝒂𝒏𝒉 𝜸𝒅
- 32. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 32 Quiz 1: A 6-m section of 150Ω lossless line is driven by a source with 𝑣𝑔 𝑡 = 5 cos 8𝜋 × 107 𝑡 − 300 (𝑉) And 𝑍𝑔 = 150Ω. If the line, which has a relative permittivity 𝜀𝑟 = 2.25 is terminated in a load 𝑍𝐿 = (150 − 𝑗50)Ω, find: a. 𝜆 on the line. Note that: 𝜆 = ൗ 𝑣𝑃 𝑓 where 𝑣𝑃 = ൗ 𝑐 𝜀𝑟 . b. The reflection coefficient at the load. c. The input impedance. d. The input voltage Vi and time-domain voltage vi(t).
- 33. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 35 8.Transmission Line Impedance andAdmittance ❖ Lossless T.L (𝛼 = 0): ▪ 𝑍𝐿 = 𝑍0: ▪ 𝑍𝐿 = 𝑗𝑋𝐿: ▪ 𝑍𝐿 = 0: ▪ 𝑍𝐿 = ∞: 𝒁(𝒙) 𝒁 𝒛 = 𝒁𝟎 𝒁𝑳 + 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 𝒁𝟎 + 𝒋𝒁𝑳𝒕𝒂𝒏 𝜷𝒅 𝒁 𝒛 = 𝒁𝟎 𝒁𝑳 + 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 𝒁𝟎 + 𝒋𝒁𝑳𝒕𝒂𝒏 𝜷𝒅 = 𝒁𝟎 𝒁 𝒛 = 𝒁𝟎 𝒋𝑿𝑳 + 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 𝒁𝟎 − 𝑿𝑳𝒕𝒂𝒏 𝜷𝒅 has imaginary part only 𝒁 𝒛 = 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 pure reactance 𝒁 𝒛 = 𝒁𝟎 𝒋𝒕𝒂𝒏 𝜷𝒅 = −𝒋𝒁𝟎𝒄𝒐𝒕𝒂𝒏 𝜷𝒅 pure reactance 𝒁 𝒛 = 𝒁𝟎 𝒁𝑳 + 𝒁𝟎𝒕𝒂𝒏𝒉 𝜸𝒅 𝒁𝟎 + 𝒁𝑳𝒕𝒂𝒏𝒉 𝜸𝒅
- 34. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 36 8.Transmission Line Impedance andAdmittance Inductance Capacitance At load ✓ Shorted-circuit T.L can be used to realize inductors or capacitors at specific frequencies → Distributed Components. ▪ 𝑍𝐿 = 0: 𝒁 𝒛 = 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 = 𝒋𝑿(𝒅) Pure reactance
- 35. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 37 8.Transmission Line Impedance andAdmittance ✓ Open-circuit T.L can be used to realize inductors or capacitors at specific frequencies → Distributed Components. ▪ 𝑍𝐿 = ∞: 𝒁 𝒛 = −𝒋𝒁𝟎𝒄𝒐𝒕𝒂𝒏 𝜷𝒅 = 𝒋𝑿(𝒅) Pure reactance Inductance Capacitance At load
- 36. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 38 8.Transmission Line Impedance andAdmittance ✓ If 𝑍𝐿 → ∞: 𝒁𝒊𝒏 = 𝟎. ✓ If 𝑍𝐿 = 0: 𝒁𝒊𝒏 → ∞. ❖ A quarter wavelength TL: 𝒁𝑳 𝒍 = 𝝀 𝟒 𝒁𝟎 𝒁𝒊𝒏 𝒁𝒊𝒏 = 𝒁𝟎 𝒁𝑳 + 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 𝒁𝟎 + 𝒋𝒁𝑳𝒕𝒂𝒏 𝜷𝒅 = 𝒁𝟎 𝟐 𝒁𝑳 ❖ Application for impedance transformation: 𝒁𝒊𝒏 = 𝒁𝟎 𝟐 𝒁𝑳 → 𝒁𝟎 = 𝒁𝒊𝒏𝒁𝑳
- 37. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Summary 2 39 𝚪𝑳 = 𝒁𝑳 − 𝒁𝟎 𝒁𝑳 + 𝒁𝟎 𝚪 𝒛 = 𝒍 − 𝒅 = 𝚪𝑳𝒆−𝟐𝜸𝒅 𝒁 𝒛 = 𝒁𝟎 𝟏 + 𝚪 𝒛 𝟏 − 𝚪 𝒛 𝒁𝒊𝒏 = 𝒁𝟎 𝟐 𝒁𝑳 → 𝒁𝟎 = 𝒁𝒊𝒏𝒁𝑳 A quarter wavelength TL
- 38. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Q&A 40
- 39. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Quiz 2 41
- 40. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 42 8.Transmission Line Impedance andAdmittance Quiz 3: The open-circuit and short-circuit impedances measured at the input terminal of a very low-loss TL of length 1.5m which is less than a quarter wavelength, are respectively -54.6j (Ω) and 103j (Ω) a. Find Z0 and 𝛾 of the line. b. Without changing the frequency, find the input impedance of a short-circuited TL that is twice the given length. c. How long should the short-circuited TL be in order to appear as an open circuit at the input terminals? 𝒁 𝒛 = 𝒁𝟎 𝒁𝑳 + 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 𝒁𝟎 + 𝒋𝒁𝑳𝒕𝒂𝒏 𝜷𝒅 𝒁 𝒛 = 𝒋𝒁𝟎𝒕𝒂𝒏 𝜷𝒅 𝒁 𝒛 = 𝒁𝟎 𝒋𝒕𝒂𝒏 𝜷𝒅
- 41. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 44 8.Transmission Line Impedance andAdmittance Quiz 4: A voltage generator with 𝑣𝑔 𝑡 = 5 cos 2𝜋 × 109 𝑡 (𝑉) and internal impedance is 𝑍𝑔 = 50Ω is connected to a 50Ω lossless T.L. The line length is 5cm and the line is terminated in a load with impedance 𝑍𝐿 = 100 − 𝑗100Ω. Determine: a. Reflection coefficient at load Γ𝐿? (phasor form) b. 𝑍𝑖𝑛 at the input of the T.L. (vp=c=3.10^8 m/s , complex form) c. The input voltage 𝑣𝑖 𝑡 and input current 𝑖𝑖 𝑡 ? (amplitude, frequency, phase)
- 42. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 46 9. PowerTransmission onTransmission Lines ❖ Steps to find 𝑽𝟎 + and 𝑽𝟎 − : 1. 𝚪𝑳 = 𝒁𝑳−𝒁𝟎 𝒁𝑳+𝒁𝟎 2. 𝚪𝒊𝒏 = 𝚪𝑳𝒆−𝟐𝜸𝒍 3. 𝒁𝒊𝒏 = 𝒁𝟎 𝟏+𝚪𝒊𝒏 𝟏−𝚪𝒊𝒏 𝒁𝑳 𝒍 = 𝝀 𝟒 𝒁𝟎 𝒁𝒊𝒏 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜸𝒛 + 𝑰𝟎 − 𝒆+𝜸𝒛 4. 𝑽𝒊𝒏 = 𝑽𝑺 𝒁𝒊𝒏 𝒁𝒊𝒏+𝒁𝑺 5. 𝑽𝒊𝒏 = 𝑽𝟎 + + 𝑽𝟎 − = 𝑽𝟎 + 𝟏 + 𝚪𝒊𝒏 6. 𝑽𝟎 + = 𝑽𝒊𝒏 𝟏+𝚪𝒊𝒏 𝑽𝟎 − = 𝚪𝒊𝒏𝑽𝟎 + ❖ If 𝒁𝑳 = 𝒁𝟎: 𝑽𝟎 + = ൗ 𝑽𝑺 𝟐 𝒘𝒉𝒆𝒏 𝒁𝒔 = 𝒁𝟎
- 43. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 47 9. PowerTransmission onTransmission Lines 𝒁𝑳 𝒁𝑺 𝑽𝑺 𝑰(𝒍) 𝑽(𝒍) 𝒛 𝒍 𝟎 𝑃𝑖𝑛𝑐 𝑃𝑡 𝑃𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 ❖ The time average power flows along a transmission line: 𝑷𝒕 = 𝟏 𝟐 𝕽𝒆 𝑽 𝒛 𝑰∗ (𝒛) = 𝟏 𝟐𝒁𝟎 𝕽𝒆 𝑽𝟎 + 𝒆−𝜶𝒛𝒆−𝒋𝜷𝒛 + 𝑽𝟎 − 𝒆𝜶𝒛𝒆𝒋𝜷𝒛 𝑽𝟎 +∗ 𝒆−𝜶𝒛𝒆𝒋𝜷𝒛 − 𝑽𝟎 −∗ 𝒆𝜶𝒛𝒆−𝒋𝜷𝒛 = 𝟏 𝟐𝒁𝟎 𝕽𝒆 𝑽𝟎 + 𝟐 𝒆−𝟐𝜶𝒛 − 𝑽𝟎 + 𝑽𝟎 −∗ 𝒆−𝒋𝟐𝜷𝒛 + 𝑽𝟎 +∗ 𝑽𝟎 − 𝒆𝒋𝟐𝜷𝒛 − 𝑽𝟎 − 𝟐 𝒆𝟐𝜶𝒛 = 𝟏 𝟐𝒁𝟎 𝑽𝟎 + 𝟐 𝒆−𝟐𝜶𝒛 − 𝑽𝟎 − 𝟐 𝒆𝟐𝜶𝒛 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒛 𝟏 − 𝚪𝒛 𝟐 = 𝑷𝒊𝒏𝒄 − 𝑷𝒓𝒆𝒇𝒍
- 44. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 48 9. PowerTransmission onTransmission Lines 𝒁𝑳 𝒛 𝒍 𝟎 𝑃𝑖𝑛𝑐 𝑃𝑡 𝑃𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 ❖ The time average absorbed by load: 𝑷𝒕 = 𝟏 𝟐 𝕽𝒆 𝑽𝑳𝑰𝑳 ∗ = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝟏 − 𝚪𝑳 𝟐 = 𝑷𝒊𝒏𝒄 − 𝑷𝒓𝒆𝒇𝒍 𝒁𝑳 𝒛 𝒍 𝟎 𝑷𝒊𝒏𝒄,𝟎 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝑷𝒊𝒏𝒄,𝒍 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝑷𝒓𝒆𝒇𝒍,𝒍 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝚪𝑳 𝟐 𝑷𝒓𝒆𝒇𝒍,𝟎 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟒𝜶𝒍 𝚪𝑳 𝟐 Power Flow: 𝑹𝒆𝒕𝒖𝒓𝒏 𝑳𝒐𝒔𝒔 = −𝟐𝟎𝒍𝒐𝒈𝟏𝟎 𝜞
- 45. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Summary 3 49 𝚪𝑳 = 𝒁𝑳 − 𝒁𝟎 𝒁𝑳 + 𝒁𝟎 𝚪 𝒛 = 𝒍 − 𝒅 = 𝚪𝑳𝒆−𝟐𝜸𝒅 𝒁 𝒛 = 𝒁𝟎 𝟏 + 𝚪 𝒛 𝟏 − 𝚪 𝒛 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜶𝒛𝒆−𝒋𝜷𝒛 + 𝑽𝟎 − 𝒆+𝜶𝒛𝒆+𝒋𝜷𝒛 𝑰 𝒛 = 𝑰𝟎 + 𝒆−𝜶𝒛 𝒆−𝒋𝜷𝒛 + 𝑰𝟎 − 𝒆+𝜶𝒛 𝒆+𝒋𝜷𝒛 ▪ Matched TL: 𝑍𝐿 = 𝑍0 → 𝚪 = 𝟎 ▪ Short circuit TL: 𝑍𝐿 = 0 → 𝚪 = −𝟏 ▪ Open circuit TL: 𝑍𝐿 = ∞ → 𝚪 = 𝟏 𝑽𝟎 + = 𝑽𝒊𝒏 𝟏 + 𝚪𝒊𝒏 , 𝑽𝟎 − = 𝚪𝒊𝒏𝑽𝟎 + 𝚪𝒊𝒏 = 𝚪𝑳𝒆−𝟐𝜸𝒍 𝑽𝒊𝒏 = 𝑽𝑺 𝒁𝒊𝒏 𝒁𝒊𝒏 + 𝒁𝑺 𝒁𝑳 𝒛 𝒍 𝟎 𝑷𝒊𝒏𝒄,𝟎 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝑷𝒊𝒏𝒄,𝒍 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝑷𝒓𝒆𝒇𝒍,𝒍 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝚪𝑳 𝟐 𝑷𝒓𝒆𝒇𝒍,𝟎 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟒𝜶𝒍 𝚪𝑳 𝟐 𝑷𝒕 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝟏 − 𝚪𝑳 𝟐 𝒁𝒊𝒏 = 𝒁𝟎 𝟏 + 𝚪𝒊𝒏 𝟏 − 𝚪𝒊𝒏 𝒁(𝒙) 𝑹𝒆𝒕𝒖𝒓𝒏 𝑳𝒐𝒔𝒔 = −𝟐𝟎𝒍𝒐𝒈𝟏𝟎 𝜞
- 46. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 50 10. StandingWave and StandingWave Ratio 𝒁(𝒙) 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛 + 𝑽𝟎 − 𝒆+𝜸𝒛 = 𝑽𝟎 + 𝒆−𝜸𝒛(𝟏 + 𝑽𝟎 − 𝑽𝟎 + 𝒆+𝟐𝜸𝒛) = 𝑽𝟎 + 𝒆−𝜸𝒛 (𝟏 + 𝚪 𝒛 ) ❖ If 𝛼 = 0: 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝒋𝜷𝒛 𝟏 + 𝚪 𝒛 → 𝑽 𝒛 = 𝑽𝟎 + 𝟏 + 𝚪 𝒛 ❖ Then: 𝑽 𝒛 𝒎𝒂𝒙 = 𝑽𝟎 + 𝟏 + 𝚪𝑳 𝒘𝒉𝒆𝒏 𝚪 𝒛 = 𝚪𝑳 𝑽 𝒛 𝒎𝒊𝒏 = 𝑽𝟎 + 𝟏 − 𝚪𝑳 𝒘𝒉𝒆𝒏 𝚪 𝒛 = − 𝚪𝑳 𝚪 𝒛 = 𝒍 − 𝒅 = 𝚪𝑳𝒆−𝟐𝜸𝒅 𝑽𝑺𝑾𝑹 = 𝑽 𝒛 𝒎𝒂𝒙 𝑽 𝒛 𝒎𝒊𝒏 = 𝟏 + 𝚪𝑳 𝟏 − 𝚪𝑳
- 47. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 51 10. StandingWave and StandingWave Ratio 𝒁(𝒙) ❖ We have: 𝑽 𝒛 = 𝑽𝟎 + 𝒆−𝒋𝜷𝒛 𝟏 + 𝚪 𝒛 𝚪 𝒛 = 𝒍 − 𝒅 = 𝚪𝑳𝒆−𝟐𝜷𝒅 = 𝚪𝑳 𝒆𝒋𝜽𝒓𝒆−𝟐𝒋𝜷𝒅 where: ❖ Then: 𝑽 𝒛 = 𝑽𝟎 + 𝟏 + 𝚪 𝒛 = 𝑽𝟎 + 𝟏 + 𝚪𝑳 𝟐 + 𝟐 𝚪𝑳 𝒄𝒐𝒔 𝟐𝜷𝒅 − 𝜽𝒓 ൗ 𝟏 𝟐 ❖ Matched TL: 𝑍𝐿 = 𝑍0 → 𝚪 = 𝟎 ❖ Short circuit TL: 𝑍𝐿 = 0 → 𝚪 = −𝟏 ❖ Open circuit TL: 𝑍𝐿 = ∞ → 𝚪 = 𝟏
- 48. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 52 10. StandingWave and StandingWave Ratio 𝑽 𝒛 = 𝑽𝟎 + 𝟏 + 𝚪𝑳 𝟐 + 𝟐 𝚪𝑳 𝒄𝒐𝒔 𝟐𝜷𝒅 − 𝜽𝒓 ൗ 𝟏 𝟐 ❖ 𝑽 𝒛 = 𝑽 𝒛 𝒎𝒊𝒏 = 𝑽𝟎 + 𝟏 − 𝚪𝑳 𝑤ℎ𝑒𝑛: 𝑐𝑜𝑠 2𝛽𝑑 − 𝜃𝑟 = −1 2𝛽𝑑 − 𝜃𝑟 = 2𝑛 + 1 𝜋 ❖ 𝑽 𝒛 = 𝑽 𝒛 𝒎𝒂𝒙 = 𝑽𝟎 + 𝟏 + 𝚪𝑳 𝑤ℎ𝑒𝑛: 𝑐𝑜𝑠 2𝛽𝑑 − 𝜃𝑟 = 1 2𝛽𝑑 − 𝜃𝑟 = 2𝑛𝜋 https://www.youtube.com/watch?v=yCZ1zFPvrIc
- 49. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 53 10. StandingWave and StandingWave Ratio Example 2: Measurements VSWR with a Z=50 slotted line terminated in an unknown load impedance is found to be 3.0. The distance between successive voltage minima is 30cm and the first minimum is located at 12cm from the load. Determine: a. The reflection coefficient Γ. b. The load impedance 𝑍𝐿
- 50. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 54
- 51. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Summary 4 56 𝚪𝑳 = 𝒁𝑳 − 𝒁𝟎 𝒁𝑳 + 𝒁𝟎 𝚪 𝒛 = 𝒍 − 𝒅 = 𝚪𝑳𝒆−𝟐𝜸𝒅 𝒁 𝒛 = 𝒁𝟎 𝟏 + 𝚪 𝒛 𝟏 − 𝚪 𝒛 𝑽𝑺𝑾𝑹 = 𝑽 𝒛 𝒎𝒂𝒙 𝑽 𝒛 𝒎𝒊𝒏 = 𝟏 + 𝚪𝑳 𝟏 − 𝚪𝑳 𝑽𝟎 + = 𝑽𝒊𝒏 𝟏 + 𝚪𝒊𝒏 , 𝑽𝟎 − = 𝚪𝒊𝒏𝑽𝟎 + 𝑽𝒊𝒏 = 𝑽𝑺 𝒁𝒊𝒏 𝒁𝒊𝒏 + 𝒁𝑺 𝒁𝑳 𝒛 𝒍 𝟎 𝑷𝒊𝒏𝒄,𝟎 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝑷𝒊𝒏𝒄,𝒍 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝑷𝒓𝒆𝒇𝒍,𝒍 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝚪𝑳 𝟐 𝑷𝒓𝒆𝒇𝒍,𝟎 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟒𝜶𝒍 𝚪𝑳 𝟐 𝑷𝒕 = 𝑽𝟎 + 𝟐 𝟐𝒁𝟎 𝒆−𝟐𝜶𝒍 𝟏 − 𝚪𝑳 𝟐 𝒁(𝒙) ❖ 𝑽 𝒛 = 𝑽 𝒛 𝒎𝒊𝒏 = 𝑽𝟎 + 𝟏 − 𝚪𝑳 𝑤ℎ𝑒𝑛: 𝑐𝑜𝑠 2𝛽𝑑 − 𝜃𝑟 = −1 2𝛽𝑑 − 𝜃𝑟 = 2𝑛 + 1 𝜋 ❖ 𝑽 𝒛 = 𝑽 𝒛 𝒎𝒂𝒙 = 𝑽𝟎 + 𝟏 + 𝚪𝑳 𝑤ℎ𝑒𝑛: 𝑐𝑜𝑠 2𝛽𝑑 − 𝜃𝑟 = 1 2𝛽𝑑 − 𝜃𝑟 = 2𝑛𝜋 𝑽 𝒛 = 𝑽𝟎 + 𝟏 + 𝚪𝑳 𝟐 + 𝟐 𝚪𝑳 𝒄𝒐𝒔 𝟐𝜷𝒅 − 𝜽𝒓 ൗ 𝟏 𝟐
- 52. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Q&A 57
- 53. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 58 Exercises Exercise 1: Two half-wave dipole antennas, each with impedance of 75Ω are connected in parallel through a pair of T.L. and the combination is connected to a feed T.L. as shown in the following figure. All lines are 50Ω lossless. a. Calculate 𝑍𝑖𝑛1 b. Calculate 𝑍𝑖𝑛 of the feed line.
- 54. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 60 Exercises Exercise 2: A 50Ω lossless line of length 𝑙 = 0.15𝜆 connects a 300MHz generator with 𝑉 𝑔 = 300𝑉 and 𝑍𝑔 = 50Ω to a load 𝑍𝐿 = 75Ω. a. Compute 𝑍𝑖𝑛 b. Compute 𝑉𝑖 and 𝐼𝑖. c. Compute the time-average power delivered to the line, 𝑃𝑖𝑛 = 1 2 ℝ𝑒 𝑉𝑖𝐼𝑖 . d. Compute 𝑉𝐿 , 𝐼𝐿 and the time-average power delivered to the load, 𝑃𝐿 = 1 2 ℝ𝑒 𝑉𝐿𝐼𝑙 . e. Compute the time-average power delivered by the generator and time-average power dissipated by in 𝑍𝑔
- 55. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering 70
- 56. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Q&A 75
- 57. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Câu 1 76
- 58. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Câu 2 77
- 59. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Câu 3 78 Câu 1: Để đo trở kháng của tải, người ta nối tải với đường dây truyền sóng không suy hao 50 như hình vẽ. Sau khi cấp tín hiệu cao tần ở tần số 3GHz, trên đường dây xuất hiện sóng đứng với điện áp hiệu dụng tại bụng sóng và nút sóng là : max 8 V V = , min 2 V V = . Bụng sóng gần tải nhất với khoảng cách 1 bs d cm = . Khoảng cách giữa bụng sóng và nút sóng liền kề là cm 2 . a. Tính vận tốc truyền sóng của đường dây truyền sóng b. Tính tỉ số sóng đứng điện áp VSWR, và dùng đồ Smith để tính trở kháng tải và hệ số phản xạ tại tải. c. Khoảng cách từ tải đến nguồn là 12cm. Tính và viết biểu thức điện áp tổng trên tải và điện áp nguồn Vs theo thời gian biết điện áp nguồn có pha là 0o . Nếu đường dây tuyền sóng ở hình trên có hệ số tổn hao là 5 dB/m, điện áp nguồn và trở kháng tải là không đổi, hãy tính công suất tại ngõ vào của đường dây và công suất trên tải theo dBm. Z0=50 ZL RS=50 VS L = 12cm
- 60. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Câu 4 79 Câu 1:(2.5 điểm) Cho đường dây truyền sóng (xem như không tổn hao) có chiều dài 2 . 1 = l như hình vẽ; trở kháng đặc tính 50 0 = R . Nguồn cấp có trở kháng nội 50. Dùng đồ thị Smith xác định: a. Vị trí điểm tải L Z trên đồ thị Smith; hệ số phản tại tải L . Tỉ số sóng đứng điện áp VSWR. b. Quỹ tích trở kháng đường dây và hệ số phản xạ khi đi từ tải về nguồn. c. Trở kháng IN Z và hệ số phản xạ tại đầu vào đường dây IN . d. Vị trí bụng sóng, nút sóng (điện áp) gần tải nhất. e. Xác định số bụng sóng và số nút sóng (điện áp) trên toàn bộ đường dây.
- 61. HCMUT / 2020 DungTrinh, PhD Dept. of Telecoms Engineering Câu 5 80